Problem: $\sum\limits_{k=1}^{800 }{{(-8k + 109)}}=$
Solution: What is the question asking for? The question is asking for the sum of the values of $-8k + 109$ from $k = 1$ to $k = 800 $ : $(-8 \cdot 1 + 109) + (-8 \cdot 2 + 109) +... + (-8\cdot {800} +109)$ The series is arithmetic because the formula $-8k + 109$ is a linear function of $k$. Formula for arithmetic series The sum $S_n$ of a finite arithmetic series is $S_n = \dfrac {\left(a_1 + a_n \right)}{2} \cdot n$ where $a_1$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. What do we need to use the formula? The number of terms $(n = {800})$ is the upper limit of the sigma notation. We need to find $a_1$ (the first term) and $a_{800}$ (the last term). Step 1: Find $a_1$ and $a_{800}$ (the first and the last term) $a_1 = -8(1) + 109 = {101}$ $a_{800} = -8(800) + 109 = {-6291}$ Step 2: Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac {\left(a_1 + a_n \right)}{2} \cdot n \\\\ S_{{800}}&= \dfrac {\left({101} + ({-6291}) \right)}{2} \cdot {800} \\\\ S_{{800}} &= -3095 \left(800\right) \\\\ S_{{800}} &= -2{,}476{,}000\end{aligned}$ The answer $ -2{,}476{,}000 $